#Find the largest palindrome made from the product of two 3-digit numbers.
from sys import exit
rev=0
print ' let us start'
for i in range(999,100,-1):
for j in range(999,100,-1):
product = i*j
temp = product
rev1 = str(product)[::-1]
a = rev1
if temp == a:
print ' is a palindrome'
if a > rev:
rev = a
c = temp
h = i
y = j
print '%r*%r=%r,which is the highest palindrome %r' % ( h, y, c, rev)
print a
print rev1
print temp
print 'over'
输出:我使用sublime text2作为编辑器
let us start
Traceback (most recent call last):
File "palindrome.py", line 19, in <module>
print '%r*%r=%r,which is the hghest palindrome %r' % ( h, y, c, l)
NameError: name 'h' is not defined
答案 0 :(得分:0)
n == a
永远不会是True
,因为k = str(t)[::-1]
(即a
)是一个字符串,而t = i*j
即n
是一个整数。尝试:
a = int(k)
答案 1 :(得分:0)
这个问题已经回答了,这里有点优化:
>>> largest, loopmax = -1, -1
>>> largest_t = (0,0)
>>> for i in range(999,100,-1):
for j in range(999,100,-1):
p = i*j
n = str(p)
if n == n[::-1]:
loopmax = p
break # no need to loop further as j decreases and can never be any greater
if loopmax > largest:
largest = loopmax
largest_t = (i,j)
if i<largest_t[1]:
break # further looping will generate only lower palindromes.
>>> largest
906609
>>> largest_t
(993, 913)
答案 2 :(得分:0)
两个3位数字的产品不到一百万,因此蛮力方法有效,找到最大(数字)产品也是回文:
all3digit = range(100, 1000) # all 3-digit numbers
products = (x*y for x in all3digit for y in all3digit)
def ispalindrome(p):
s = str(p)
return s == s[::-1]
print(max(filter(ispalindrome, products))) # -> 906609